1031. Hello World for U (20)
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Given any string of N (>=5) characters, you are asked to form the characters into the shape of U. For example, "helloworld" can be printed as:
h d e l l r lowo
That is, the characters must be printed in the original order, starting top-down from the left vertical line with n1 characters, then left to right along the bottom line with n2 characters, and finally bottom-up along the vertical line with n3 characters. And more, we would like U to be as squared as possible -- that is, it must be satisfied that n1 = n3 = max { k| k <= n2 for all 3 <= n2 <= N } with n1 + n2 + n3 - 2 = N.
Input Specification:
Each input file contains one test case. Each case contains one string with no less than 5 and no more than 80 characters in a line. The string contains no white space.
Output Specification:
For each test case, print the input string in the shape of U as specified in the description.
Sample Input:
helloworld!
Sample Output:
h ! e d l l lowor
题目大意:把字符串的输出按U形样式输出。
解题思路:可根据公式n1=n3=max{k|k<=n2 for all 3<=n2<=N}和n1+n2+n3-2=N,对n1与n2进行遍历,先求解出n1和n2的值。
#include<iostream> #include<cstdio> #include<cstring> using namespace std; int main(){ char s1[81],s2[81][81]; scanf("%s",s1); int length=strlen(s1); int i,j; int n = length+2; int flag=0; for(i=1;i<=length;i++){ for(j=1;j<=i;j++){ if((i+j*2)%n == 0){ flag = 1; break; } } if(flag==1)break; } int n2=i-2; int n3,n1; n3=n1=j; length-=1; for(i=0;i<n1-1;i++){ printf("%c",s1[i]); for(j=0;j<n2;j++){ printf(" "); } printf("%c\n",s1[length-i]); } for(i=n1-1;i<=length-n1+1;i++){ printf("%c",s1[i]); } printf("\n"); return 0; }
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